Discriminate convex vs concave vertices within each polygon feature

teakplantation

Using the calculate geometry tool, one can calculate the number of vertices for each polygon in a feature class (https://pro.arcgis.com/en/pro-app/latest/tool-reference/data-management/calculate-geometry-attributes.htm).

A vertex can be concave or convex, depending on whether the inner angle at the vertex is larger or smaller than 180 degrees.

Does anybody know of a tool which can calculate how many vertices per polygon are convex and how many are concave?

 

JohannesLindner

Starting from Dan's suggestion of convexHull(), which seemed good:

with arcpy.da.UpdateCursor("TestPolygons", ["SHAPE@", "IntegerField"]) as cursor:
    for shp, n in cursor:
        n = len(shp[0]) - len(shp.convexHull()[0])  # only for singlepart polygons
        cursor.updateRow([shp, n])

 

This works well for simple geometries:

 

But it starts to break down for more complex geometries, where convex vertices are not always on the convex hull:

 

I was ready to start fussing with angles between polylines when I realized that you can quite simply say whether a vertex is concave or convex: cretae a triangle of the vertex and its neighbours. If that triangle is inside the polygon, the vertex is convex, else it's concave.

 

def get_concave_vertex_count(polygon):
    n = 0
    for part in polygon:
        for i, vertex in enumerate(part):
            prev_vertex = part[i-1]
            try:
                next_vertex = part[i+1]
            except IndexError:
                next_vertex = part[0]
            triangle = arcpy.Polygon(arcpy.Array([prev_vertex, vertex, next_vertex]))
            n += triangle.touches(polygon)  # touch: only boundaries intersect
    return n


with arcpy.da.UpdateCursor("PolygonLayer", ["SHAPE@", "ConcaveVertices"]) as cursor:
    for shp, n in cursor:
        n = get_concave_vertex_count(shp)
        cursor.updateRow([shp, n])